Question

use logarithmic differentiation to find the derivative: y=(lnx)^cosx

Answer 1

that doesn't sound right. It should be ln (x^cos x) ?

Answer 2

it is 100% y=(lnx)^cosx

Answer 3

i got 1? like xcosx/xcosx

Answer 4

no, \[\int\limits 1 dx = x+C\]

Answer 5

ohh I am not that advanced! I dont know what that means! I had to use the logarithmic differentiation to find my answer!

Answer 6

so y=cosx(lnx)
y=cosx(1/x)
lny=ln(cosx/x)
y'= (x/cosx)y
y=x(cosx/x)/cosx
y=xcosx/xcosx

Answer 7

????????
okay your solution is ridiculously wrong, I could see at least 2 mistakes upon sight.
First, you cannot differentiate ln x inside bracket first;
you want to find dy/dx, however at line 3 and 4 the left side is 1; Differentiating that would leave left side = 0;
on 5th line and 6th line you even got division wrong.

Answer 8

y=(lnx)^cosx
If this is the true equation it does require substitution and chain rule, but I don't see how log differentiation would be used.

Answer 9

I have it all sorted out, thanks anyways ??? for sort of helping, but not really. Thats the true equation from the text book, with the directions to solve using log differentiation.